Integer solutions to $ n^2 + 1 = 2 \times 5^m$

By factoring the given equation over the Gaussian integers, we are looking for solutions of: $$a_m=\operatorname{Re}\left((1\pm i)(2+i)^m\right)=\pm 1.\tag{1}$$ It is straighforward to check that $(1)$ is fullfilled by $m=0,m=1$ and $m=2$.

In order that $(1)$ holds, $\frac{\pi}{4}+m\arctan\frac{1}{2}\pmod{2\pi}$ must be close to $0,\pm\frac{\pi}{2}$ or $\pi$.

That happens for $m=15,22,29,36,\ldots$ and many others $m$ such that: $$\left\{m\cdot\frac{1}{\pi}\arctan\frac{1}{2}\right\}\text{ is close to }\frac{1}{4}\text{ or } 0.\tag{2}$$ where $\{\cdot\}$ stands for the fractional part. So we have that other solutions of $(1)$ may occur for $m=q$ or $m=q/4$, where $\frac{p}{q}$ is a convergent of the continued fraction of $\alpha=\frac{1}{\pi}\arctan\frac{1}{2}$ and $4\mid q$ if needed.

The first of such good candidates is $m=83$: in such a case the absolute value of the ratio between the real part and the imaginary part of $(1+i)(2+i)^m$ is something like $0.00175$, but the real part is still a $27$-digits number.

It is reasonable to conjecture that no other solutions exist apart from the trivial ones; however, that looks hard to prove (at least to me) since the continued fraction of $\alpha$ does not have a good structure (like for the golden ratio, the constant $e$ or some values of Bessel's functions).

An alternative approach (but not so much) may be to prove by arithmetical arguments that the only solutions of the Pell equations $$ n^2-2N^2=1,\qquad n^2-10 N^2=1\tag{3}$$ with $N=5^m$ are the trivial ones. That problem depends on the convergents of $\sqrt{2}$ and $\sqrt{10}$ that are way nicer. It is probably not to difficult to derive an explicit formula for the $5$-adic height of a lambda number $P_n$ and deduce that there are a finite number of $N$s of the form $5^m$ that may fulfill $(3)$.